10056 - What is the Probability ? 解题报告
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题目链接:
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=997
N个人轮流掷筛子,赢的概率为P,求第i个人赢的概率是多少。
可以求出第i个人赢的概率为P(1-P)^(i-1) * [1 + (1-P)^N + (1-P)^2N + ...] ,后边是一个等比数列,可以化简为P(1-P)^(i-1) *(1-(1-P)^N^2)/(1-(1-P)^N),其中飘红的部分趋近于0,因此最终的解为P(1-P)^(i-1) /(1-(1-P)^N),有一点需要注意:由于分母不能为0,所以此公式对于P为0(或非常小的一个值)的情况是不适用的,P为0时,任何人赢的概率都为0
代码如下:
#include<cmath>
#include<iomanip>
#include<iostream>
using namespace std;
#define eps 1e-9
int main()
{
int s;
cin>>s;
int n, player;
double p;
while(s--)
{
cin>>n>>p>>player;
double a = 1;
a = pow(1-p, n);
double q, ans;
q = p*pow(1-p, player-1);
if(p < eps && p > -eps) //对0特殊处理
ans = 0;
else
ans = q/(1-a);
cout<<fixed<<setprecision(4)<<ans<<endl;
}
return 0;
}
换一种方式:第一轮掷筛子的过程中,第i个人赢的概率为P(1-P)^(i-1),那么第一轮没有人赢的概率为(1-P)^N,后续每一轮中每个人赢的比例是固定的,那么第i个人赢的概率为
P(1-P)^(i-1)/(1-(1-P)^N),跟上边推出的公式一致
测试case:
3
2 0.00000000001 1
2 0.166666 1
2 0.166666 2
结果为:
0.0000
0.5455
0.4545
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